Remember the old puzzle?

"Can you take just one letter out of an odd number and make it even?"

Note the clever wording. It just says make it even; not "make it an even number"...

So here is my version:

Find the smallest odd integers and remove "n" letters. The remaining letters must be an even number. The "n" is from 2 to 9. (I could not find a solution to just taking away 1 letter.) The integers must be less than 100. Once the letters are removed you cannot rearrange the remaining letters.

No partial answers please. Need 8 numbers.

If you can find a solution for n=1, please report that.

  • $\begingroup$ I see everyone has answered with positive integers. What's your definition of "smallest"? $\endgroup$ Apr 6, 2021 at 21:32

2 Answers 2


They are

2 five to iv (roman numerals)
3 twenty-one to twenty
4 fifteen to ten
5 thirteen to ten
6 seventeen to ten
7 twenty-five to ten
8 twenty-three to ten
9 seventy-seven to ten

Thanks to @LukasRotter for reminding me it was odd to even

@Amoz and @Stiv suggested some rather clever ways for a solution to n=1

iii to ii (@amoz)
ONe to Ne (atomic number 10... even!) (@stiv)

Also, the solution to the "old joke" mentioned is

seven to even

  • $\begingroup$ Base numbers have to be odd. So just +1 for your #3, #4 and #7 $\endgroup$ Apr 5, 2021 at 13:46
  • $\begingroup$ @LukasRotter oh doh, missed that requirement. thanks! $\endgroup$
    – SteveV
    Apr 5, 2021 at 13:47
  • 2
    $\begingroup$ If five to iv is legitimate, is iii to ii valid? (removing one) $\endgroup$
    – Amoz
    Apr 5, 2021 at 20:57
  • $\begingroup$ @Amoz looks like you found a 1 removal! $\endgroup$
    – SteveV
    Apr 5, 2021 at 21:44
  • 1
    $\begingroup$ @Stiv you have periodic table prowess today! If you don't mind, I'll add yours and Amoz's cleverness to the answer $\endgroup$
    – SteveV
    Apr 6, 2021 at 14:26

As an alternative to @SteveV's strategy, there is another way to approach this puzzle, by considering:

numbers whose names can be reduced by removing letters to leave the letter 'E' or the digraph 'EE', which are the hexadecimal equivalents of the decimal values '14' and '238', respectively, i.e. even numbers! (There's nothing in the original question to preclude using base systems other than decimal, after all...)

Using this method, it is possible to use even smaller odd numbers than in that answer, right across the board:

Removing 2 letters: ONE → E (14)
3: THREE → EE (238)
4: THREE → E (14)
5: ELEVEN → E (14)
6: THIRTEEN → EE (238)
7: THIRTEEN → E (14)
8: SEVENTEEN → E (14)
9: TWENTY THREE → EE (238)

(NB There is no point considering the letters A, B, C or D, which are valued 10-13 in hexadecimal, since there are no English odd numbers less than 100 - a stipulation of the original puzzle - that contain these letters... You'll need to count as high as one-hundred-and-one, one-billion-and-one, or one-octillion-and-one before you find them! And while we could consider combinations involving the letter 'F' (15) also, the fact that it is rarer than the letter 'E' in the names of small odd numbers actually means there is always a smaller solution using 'E's alone...)

Note that it is impossible to use this same technique to find a solution requiring the removal of just one letter...

Unless we 'up our game' even further and switch from using hexadecimal to using a higher base, like base-24! Then we can legally do:

Removing 1 letter: ONE → NE (566 in base-24)


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